Compound Interest Questions and Answers

Answer: D

Compound interest needs both the principal and the rate, but the question supplies only the final amount.

\[ P + \frac{P \times R \times 2}{100} = 1344 \]

That is one equation in two unknowns, so infinitely many pairs fit. For example:

• \( P = 1200 \) at 6% gives 1344, and CI would be Rs. 148.32
• \( P = 1120 \) at 10% also gives 1344, and CI would be Rs. 235.20

Since the two produce different answers, the data are insufficient. Any of these questions that fixes only the amount, and not the principal or rate separately, is underdetermined.

Answer: C

The two sources of growth act in the same year, so combine them into a single annual rate.

\[ 4\% + 1\% = 5\% \text{ per annum} \]

Population growth compounds, so over two years:

\[ (1.05)^{2} = 1.1025 \]

\[ \text{increase} = 10.25\% \]

Simply doubling the 5% would give 10%; the extra 0.25% is the second year's growth on the first year's addition.

Answer: C

\[ A = P\left(1 + \frac{R}{100}\right)^{n}, \qquad \text{CI} = A - P, \qquad \text{SI} = \frac{PRn}{100} \]

The rates differ by year, so multiply the two factors rather than raising one to a power.

\[ A = 5000 \times 1.04 \times 1.05 \]

\[ = 5000 \times 1.092 = \text{Rs. } 5460 \]

The question asks for the amount, so the Rs. 460 of interest is only an intermediate value - and it is offered as the first distractor.

Answer: B

\[ A = P\left(1 + \frac{R}{100}\right)^{n}, \qquad \text{CI} = A - P, \qquad \text{SI} = \frac{PRn}{100} \]

\[ A = 8000 \times (1.1)^{2} = 8000 \times 1.21 = \text{Rs. } 9680 \]

The interest is Rs. 1680, of which Rs. 1600 is simple interest and Rs. 80 is the second year's 10% on the first year's Rs. 800.

Answer: B

Let the sum be Rs.P

 P{ [ 1 + 8/100]² - 1 } = 2828.80 

P(8/100)(2 + 8/100) = 2828.80 [a² - b² = (a - b) ( a + b)] P = 2828.80 / (0.08)(2.08) = 1360/0.08 = 17000 Principal + Interest = Rs. 19828.80

Answer: D

\[ A = P\left(1 + \frac{R}{100}\right)^{n}, \qquad \text{CI} = A - P, \qquad \text{SI} = \frac{PRn}{100} \]

\[ \text{CI} = 14800\left[(1.135)^{2} - 1\right] \]

\[ (1.135)^{2} = 1.288225 \]

\[ \text{CI} = 14800 \times 0.288225 = \text{Rs. } 4265.73 \approx \text{Rs. } 4266 \]

Simple interest would be \( 14800 \times 0.27 = 3996 \); the extra Rs. 270 is 13.5% of the first year's Rs. 1998.

Answer: D

Let the rate of interest be R% p.a. 

17400[1 + R/100]² = 17400 + 1783.50 

[1 + R/100]² = (17400 + 1783.50)/17400 

= 1 + 1025/10000 = 1 + 41/400 

= 441/400 = [21/20]² 

[1 + R/100] = 21/20 

R/100 = 1/20 

Therefore R = 5

Answer: C

Each year's interest is the previous year's grown by the rate, because the balance itself grows by that factor.

\[ 1 + \frac{R}{100} = \frac{1440}{1200} = 1.2 \]

\[ R = 20\% \text{ p.a.} \]

The difference of Rs. 240 between the two years is exactly 20% of the earlier Rs. 1200 - the interest earned on that interest. Neither the principal nor the first year's interest is needed.

Answer: B

Let A be the amount received at the end of the three years. 

A = 4800[1 + 10/100][1 + 20/100][1 + 25/100] 

A = (4800 * 11 * 6 * 5)/(10 * 5 * 4) 

A = Rs.7920 

So the interest = 7920 - 4800 = Rs.3120

Answer: D

Let the principal be Rs.P 

S.I at 5% p.a in 8 years on Rs.P = Rs.500

(P)(8)(5)/100 = 500

P = 1250

C.I on Rs.2P i.e., Rs.2500 at 5% p.a in two years

=2500{ [1 + 5/100]² - 1} = 2500{ 21² - 20² /20²} 

= 2500/400(441 - 400)

= 25/4(41) = 1025/4 = Rs.256.25